Question

Glass Houses Pty has found that the cost of fitting new sliding doors is normally distributed, with an average cost per door of R12 750, and a standard deviation of R1 095. A sample of 105 new windows is randomly selected. Assuming normal distribution: What is the probability that the average cost of fitting the doors in

the sample is less than R13 000? Interpret your answer.

Answer 1

Let $\bar{X}=$ the average cost of fitting the doors in the sample: $\bar{X}\sim N(\mu_{\bar{X}}, \sigma_{\bar{X}}^2)$

$$\mu_{\bar{X}}=\mu, \sigma_{\bar{X}}^2=\sigma^2/n$$

Given $\mu=12750, \sigma=1095, n=105.$

$$P(\bar{X}<13000)=P(Z<\dfrac{13000-12750}{1095/\sqrt{105}})$$

$$\approx P(Z<2.3395)\approx0.990345$$

The probability that the average cost of fitting the doors in the sample is less than R13 000 is $0.990345.$